有没有办法将C#通用字典拆分成多个字典?

时间:2010-02-01 15:19:27

标签: c# linq dictionary split

我有一个C#字典Dictionary<MyKey, MyValue>,我希望根据Dictionary<MyKey, MyValue>将其拆分为MyKey.KeyType的集合。 KeyType是一个枚举。

然后我会留下一个包含键值对的字典,其中MyKey.KeyType = 1和另一个字典MyKey.KeyType = 2,等等。

有没有一种很好的方法,比如使用Linq?

6 个答案:

答案 0 :(得分:9)

var dictionaryList = 
    myDic.GroupBy(pair => pair.Key.KeyType)
         .OrderBy(gr => gr.Key)  // sorts the resulting list by "KeyType"
         .Select(gr => gr.ToDictionary(item => item.Key, item => item.Value))
         .ToList(); // Get a list of dictionaries out of that

如果你想要一个最后用“KeyType”键入的词典字典,方法是类似的:

var dictionaryOfDictionaries = 
    myDic.GroupBy(pair => pair.Key.KeyType)
         .ToDictionary(gr => gr.Key,         // key of the outer dictionary
             gr => gr.ToDictionary(item => item.Key,  // key of inner dictionary
                                   item => item.Value)); // value

答案 1 :(得分:1)

我相信以下内容会有效吗?

dictionary
    .GroupBy(pair => pair.Key.KeyType)
    .Select(group => group.ToDictionary(pair => pair.Key, pair => pair.Value);

答案 2 :(得分:0)

所以你真的想要一个IDictionary<MyKey, IList<MyValue>>类型的变量?

答案 3 :(得分:0)

您可以使用GroupBy Linq函数:

            var dict = new Dictionary<Key, string>
                   {
                       { new Key { KeyType = KeyTypes.KeyTypeA }, "keytype A" },
                       { new Key { KeyType = KeyTypes.KeyTypeB }, "keytype B" },
                       { new Key { KeyType = KeyTypes.KeyTypeC }, "keytype C" }
                   };

        var groupedDict = dict.GroupBy(kvp => kvp.Key.KeyType);

        foreach(var item in groupedDict)
        {
            Console.WriteLine("Grouping for: {0}", item.Key);
            foreach(var d in item)
                Console.WriteLine(d.Value);
        }

答案 4 :(得分:0)

除非您只想拥有单独的集合:

Dictionary myKeyTypeColl<KeyType, Dictionary<MyKey, KeyVal>>

答案 5 :(得分:0)

Dictionary <int,string> sports;
sports=new Dictionary<int,string>();
sports.add(0,"Cricket");
sports.add(1,"Hockey");
sports.add(2,"Badminton");
sports.add(3,"Tennis");
sports.add(4,"Chess");
sports.add(5,"Football");
foreach(var spr in sports)
console.WriteLine("Keu {0} and value {1}",spr.key,spr.value);

输出:

Key 0 and value Cricket
Key 1 and value Hockey
Key 2 and value Badminton
Key 3 and value Tennis
Key 4 and value Chess
Key 5 and value Football